The state model of the knee-quadriceps can be expressed as

$$\begin{cases} \begin{align*} \dot{x}_1 &= \left[ s_0 \alpha K_m + s_v q\dfrac{s_0\alpha F_mx_1 - s_ux_2x_1}{1 + px_1 - s_vqx_2}\right] u_{ch} - s_ux_1u_{ch} - \dfrac{s_v ax_1 r_p x_4}{L_0 (1+px_1-s_vqx_2)}\\ \dot{x}_2 &= \left[ \dfrac{s_0\alpha F_m - s_ux_2}{1 + px_1 - s_vqx_2} \right]u_{ch} + \dfrac{bx_1r_px_4 - s_vax_2r_px_4}{L_0(1+px_1-s_vqx_2)}\\ \dot{x}_3 &= x_4\\ \dot{x}_4 &= \dfrac{1}{I}[x_2r_p - \lambda x_3 - \mu x_4 - mgl_c \cos{x_3}], \end{align*} \end{cases}$$

where $\textbf{x}=[x_1, \ldots, x_4]^{\mathrm T} = [K_c, F_c, \theta, \dot{\theta}]^{\mathrm T}$ is the state vector and $\textbf{u}=[u_{ch},\alpha ]^{\mathrm T}$ the control vector. The variable $\theta$ represents the knee joint angle and the variables $K_c, F_c, u_{ch}, \alpha$ represent the state variables of the quadriceps muscle model.

The present study of tension leg platform is the first commercial application of a revolutionary design of offshore production platform developed by well-known oil company. Intended for oil and gas production in water depths beyond the reach of traditional fixed structures, the tension leg platform was designed as a rectangular shaped floating platform which was connected to the ocean floor by 16 vertical steel tethers or legs, four per corner. The legs were kept in tension so that vertical movement was suppressed, while limited horizontal movement may occur.

The estimation model is:

$$\begin{align*} y(k) &= 0.590y(k − 3)+1.0598y(k − 1) − 1.0931y(k − 2) + 121.13u(k − 1)u(k− 1)u(k − 9) \\ &− 116.54u(k − 6)u(k − 6)u(k− 6) − 19.797u(k − 4)u(k − 8)u(k − 8) \\ &+ 214.04u(k − 5)u(k − 9) − 34.877u(k− 1)u(k − 1)u(k − 1) − 3.7983u(k − 1)u(k− 2)u(k − 7) \\ &− 25.04u(k − 4)u(k − 8)u(k− 11) + 165.93u(k − 2)u(k − 3)u(k − 4) \\ &− 173.85u(k − 6)u(k − 7) − 69.693u(k− 4)u(k − 12) + 203.12u(k − 5)u(k − 6)u(k − 6) \\ &+ 727.86u(k − 2)u(k − 3)u(k − 5) − 11.107u(k− 3)u(k − 10)u(k − 11) \\ &+ 11.506u(k − 6)u(k− 6)u(k − 12) − 68.607u(k − 2)u(k − 4)u(k− 6) \\ &− 366.75u(k − 3)u(k − 5)u(k − 6)− 25.696u(k − 4)u(k − 8)u(k − 12) \\ &+ 137.86u(k − 1)u(k − 2)u(k − 5)− 142.24u(k − 2)u(k − 2)u(k − 9) \\ &+ 101.44u(k− 1)u(k − 6)u(k − 9) − 9.0283u(k − 3)u(k− 3)u(k − 12) \\ &− 168.30u(k − 2)u(k − 5)u(k − 6)+ 30.295u(k − 5)u(k − 6)u(k − 8) \\ &− 0.158u(k− 1)u(k − 2)u(k − 2) − 433.21u(k − 2)u(k− 2)u(k − 4) \\ &+ 39.88u(k − 3)u(k − 8)u(k − 11)− 162.26u(k − 1)u(k − 4)u(k − 11) \\ &− 212.08u(k− 1)u(k − 1)u(k − 5) − 438.7u(k − 3)u(k− 3)u(k − 5) \\ &+ 162.15u(k − 2)u(k − 4)u(k − 11)− 3.607u(k − 4)u(k − 4)u(k − 11) \\ &+ 13.262u(k− 6)u(k − 9)u(k − 9)+448.4u(k − 3)u(k − 4)u(k− 6) \\ &− 46.475u(k − 4)u(k − 4)u(k − 9) + 119.95u(k − 1)u(k − 1)u(k − 2) + noise \:terms. \end{align*}$$

Here, the input is the wave, and the output is the pitch. Sample rate is 2.2473 Hz

Consider the following time varying stochastic bilinear system with nonlinear feedback.

$$\begin{align*} \begin{bmatrix} x_1(t+1) \\ x_2(t+1) \end{bmatrix} &= \left\{\begin{bmatrix}0.1 & 0.2 \\ 0.5 & -0.3\end{bmatrix} + \begin{bmatrix}0.36 & -0.3 \\ 0.2 & 0.42\end{bmatrix}\omega(t) \right\}\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix} \\ &+\begin{bmatrix}0.1 & 0.9 \\ 1.5 & 1.2\end{bmatrix} \begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}u(t) + \begin{bmatrix}-0.3t^2\exp{(-t)} \\ 0.4t\exp{(-t)}\end{bmatrix}u(t), \\ \begin{bmatrix}y_1(t) \\ y_2(t)\end{bmatrix} &= \begin{bmatrix} 0.7\sin{t} & -0.9 \\ 0.8 & -0.6\cos{t}\end{bmatrix} \begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}, \end{align*}$$

where

$u(t)=0.2\sin{(y_1(t) + y_2(t))} + 0.3[y_1(t)+y_2(t)].$

Consider the following time invariant stochastic bilinear system:

$$\begin{align*} \begin{bmatrix}x_1(t+1)\\x_2(t+1)\end{bmatrix} &= \left\{ \begin{bmatrix}0.2 & 0.4 \\ 0.5 & -0.3\end{bmatrix} + \begin{bmatrix}0.3 & 0.2 \\ -0.3 & 0.4\end{bmatrix}\omega(t) \right\} \begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}+ \begin{bmatrix}2 & 5 \\ 3 & 9\end{bmatrix} \begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}u(t) + \begin{bmatrix}-0.3 \\ 0.4\end{bmatrix}u(t), \\ \begin{bmatrix}y_1(t) \\ y_2(t)\end{bmatrix}& = \begin{bmatrix} 0.7 & 0.8 \\ -0.9 & -0.6\end{bmatrix} \begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}, \end{align*}$$

where

$u(t)=0.24[y_1(t) + y_2(t)] + 0.32[y_1(t-1) + y_2(t-1)]$

and $\omega(t)$ is a white noise with zero mean and variance 0.2.

The time-invariant bilinear system is given by

$$Y(t) = 1.5X(t) + 1.2X(t-1) - 0.2X(t-2) + 0.7X(t-1)Y(t-1) - 0.1X(t-2)Y(t-2) + \epsilon(t),$$

where $A=0, \alpha=0, B=\begin{bmatrix}1.5 &1.2 &-0.2\end{bmatrix}, C = \begin{bmatrix}0.7 &0 &-0.1\end{bmatrix}$. Note that $\Theta = \begin{bmatrix}B & C\end{bmatrix}^{\mathrm T}.$